A consumer group was interested in comparing the operating time of cordless toot

Question

A consumer group was interested in comparing the operating time of cordless toothbrushes manufactured by two different companies. They took a random sample of toothbrushes from each company. Each toothbrush was charged overnight and the number of hours of use before needing to be recharged was recorded. The data are shown in the table.  

Do these samples indicate that there is a difference in the mean operating times between cordless toothbrushes manufactured by the two companies? Use = 0.05

Company 1

Company 2

Sample size

Sample mean

Sample standard deviation

24. Which of the following statements is true?

A. This is a one tailed test of two dependent samples
B. This is a two tailed test of two independent samples.                 
C. This is a one tailed test of two independent samples.
D. These samples are matched.
E. None of the above.

25. The correct Null and Alternative hypotheses are

A. H0: µ1 - µ2 0 and HA: µ1 - µ2 > 0
B. H0: µ1 - µ2 0 and HA: µ1 - µ2 < 0
C. H0: µ1 - µ2 = 0 and HA: µ1 - µ2 < 0
D. H0: µ1 - µ2 = 0 and HA: µ1 - µ2 0             
E. None of the above

26. The correct value of the test statistic is

A. t = -1.489  
B. t = -4.421
C. t = 2.566
D. t = -0.488
E. None of the above

27. At = .05

A. we reject the null hypothesis.
B. we fail to reject the null hypothesis.
C. we can conclude that there is no significant difference between the mean operating times of cordless toothbrushes manufactured by the two companies.
D. Both A and C.
E. Both B and C.

Company 1

Company 2

Sample size

Sample mean

Sample standard deviation

Explanation / Answer

Solution:-

24) (B) This is a two tailed test of two independent samples.

25) (D)State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 = 2

Alternative hypothesis: 1 2  

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

S.E = sqrt[(s12/n1) + (s22/n2)]

S.E = 0.6045

DF = 30.057 (BY using online calculator)

D.F = 30

t = [ (x1 - x2) - d ] / SE

26 (A) t = - 1.489

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 40 degrees of freedom is more extreme than -1.49; that is, less than -1.49 or greater than 1.49.

Thus, the P-value = 0.14693

Interpret results. Since the P-value (0.14693) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that there is  is a difference in the mean operating times between cordless toothbrushes manufactured by the two companies.

27 (C) We can conclude that there is no significant difference between the mean operating times of cordless toothbrushes manufactured by the two companies.

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